Three normals are drawn from the point (c, 0) to the curve $y^2=x$ . One normal is always the x-axis. Find c for which the other two normals are perpendicular to each other .

The equation of parabola is $y^2=4ax$  and normal to this parabola is  $y=mx-2am-a{m}^3$
Here it is given that  $y^2=x\text{\hspace{0.17em}so\hspace{0.17em}}4a=1\text{\hspace{0.17em}or\hspace{0.17em}}a=\frac{1}{4}$ so equation of normal to the parabola (curve) becomes  $0=mc-\frac{m}{2}-\frac{m^3}{4}$
 $\Rightarrow m=0or{m}^2=4(c-\frac{1}{2})$
Here $m^2\ge 0\text{\hspace{0.17em}so\hspace{0.17em}}4(c-\frac{1}{2})\ge 0\Rightarrow c\ge \frac{1}{2}$  Thus at  $c=\frac{1}{2}\Rightarrow m^2=0\Rightarrow m=0$
Now c must be greater than zero for real values of m.
Thus for m = 0, y = 0. It means x – axis is the normal. Now we have  $m^2=4(c-\frac{1}{2})$ and we have that  $m_1{m}_2=-1$
Since other two normals are perpendicular
 $\Rightarrow -4(c-\frac{1}{2})=-1\Rightarrow c=\frac{3}{4}$